Question

Consider a particle of mass m in the one-dimensional 6 function potential

V(z) = V₀ δ(x).

Show that if Vo is negative there exists a bound state, and that the binding energy is \(mV_0^2/2h^2\).

Answer 1

In the Schrodinger equation

Integrating both sides over x from \(-\varepsilon\) to \(+\varepsilon\), where \(\varepsilon\) is an arbitrarily small positive number, we obtain

With \(\varepsilon \to 0^+\), this becomes \(\psi '(0^+) - \psi'^{0^-} = U_0\psi(0)\). For x \(\ne\) 0 the Schrodinger equation has the formal solution \(\psi (x)\sim\exp(-k|x|)\) with k positive, which gives

Thus k = -U₀/2, which requires V₀ to be negative. The energy of the bound state is then E = \(- \frac{h^2k^2}{2m}= -m V_0^2/2h^2\) and the binding energy is \(E_b = 0 - E = mV_0^2/2h^2\). The wave function of the bound state is

where the arbitrary constant A has been obtained by the normalization

\(\int \limits_{-\infty}^0\psi^2 dx + \int\limits_\infty^0\psi^2dx = 1\)